What is the solution for (x^2-9)(dy)/(dx)+6y=(x+3)^2 ?

The solution for (x^2-9)(dy)/(dx)+6y=(x+3)^2 is y=(x(x+3))/(x-3)+(c_{1}(x+3))/(x-3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2-9)(dy)/(dx)+6y=(x+3)^2

(x^{2} - 9) \frac{d y}{d x} + 6 y = (x + 3)^{2}

y = \frac{x ( x + 3 )}{x - 3} + \frac{c _{1} ( x + 3 )}{x - 3}

Solution steps

(x^{2} - 9) \frac{d y}{d x} + 6 y = (x + 3)^{2}

Solve linear ODE:

y = \frac{x ( x + 3 )}{x - 3} + \frac{c _{1} ( x + 3 )}{x - 3}

y = \frac{x ( x + 3 )}{x - 3} + \frac{c _{1} ( x + 3 )}{x - 3}

x \to 1 + lim (\frac{x - 4}{1 - x})

\frac{\partial}{\partial x} (e^{x - 1})

5 e^{3 t}

\int \frac{x}{x ^{2} + 3} d x

\frac{\partial}{\partial x} (tan (ln (x \cdot y)))

What is the solution for (x^2-9)(dy)/(dx)+6y=(x+3)^2 ?
The solution for (x^2-9)(dy)/(dx)+6y=(x+3)^2 is y=(x(x+3))/(x-3)+(c_{1}(x+3))/(x-3)